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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2009, Article ID 486375, 7 pages
doi:10.1155/2009/486375
Research Article
Superstability of Generalized Multiplicative
Functionals
Takeshi Miura,
1
Hiroyuki Takagi,
2
Makoto Tsukada,
3
and Sin-Ei Takahasi
1
1
Department of Applied Mathematics and Physics, Graduate School of Science and Engineering,
Yamagata University, Yonezawa 992-8510, Japan
2
Department of Mathematical Sciences, Faculty of Science, Shinshu University,
Matsumoto 390-8621, Japan
3
Department of Information Sciences, Toho University, Funabashi, Chiba 274-8510, Japan
Correspondence should be addressed to Takeshi Miura,
Received 2 March 2009; Accepted 20 May 2009
Recommended by Radu Precup
Let X be a set with a binary operation ◦ such that, for each x, y, z ∈ X,eitherx ◦y ◦z x◦z ◦y,
or z◦x ◦yx◦z◦y. We show the superstability of the functional equation gx ◦ygxgy.
More explicitly, if ε ≥ 0andf : X →
C satisfies |fx ◦ y − fxfy|≤ε for each x, y ∈ X,then
fx ◦ yfxfy for all x, y ∈ X,or|fx|≤1
√
1 4ε/2forallx ∈ X. In the latter case, the
constant 1
√
1 4ε/2 is the best possible.
Copyright q 2009 Takeshi Miura et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Introduction
It seems that the stability problem of functional equations had been first raised by S. M. Ulam
cf. 1, Chapter VI. “For what metric groups G is it true that an ε-automorphism of G is
necessarily near to a strict automorphism? An ε-automorphism of G means a transformation
f of G into itself such that ρfx·y,fx ·fy <εfor all x, y ∈ G.” D. H. Hyers 2 gave an
affirmative answer to the problem: if ε ≥ 0andf : E
1
→ E
2
is a mapping between two real
Banach spaces E
1
and E
2
satisfying fx y − fx − fy≤ε for all x, y ∈ E
1
, then there
exists a unique additive mapping T : E
1
→ E
2
such that fx − Tx≤ε for all x ∈ E
1
.If,
in addition, the mapping R t → ftx is continuous for each fixed x ∈ E
1
, then T is linear.
This result is called Hyers-Ulam stability of the additive Cauchy equation gx ygx
gy.J.A.Baker3, Theorem 1 considered stability of the multiplicative Cauchy equation
gxygxgy:ifε ≥ 0andf is a complex valued function on a semigroup S such that
|fxy − fxfy|≤ε for all x, y ∈ S, then f is multiplicative, or |fx|≤1
√
1 4ε /2
2 Journal of Inequalities and Applications
for all x ∈ S. This result is called superstability of the functional equation gxygxgy.
Recently, A. Najdecki 4, Theorem 1 proved the superstability of the functional equation
gxφy gxgy:ifε ≥ 0, f is a real or complex valued functional from a commutative
semigroup X, ◦, and φ is a mapping from X into itself such that |fx ◦φy −fxfy|≤ε
for all x, y ∈ X, then fx ◦ φy fxfy
holds for all x, y ∈ X,orf is bounded.
In this paper, we show that superstability of the functional equation gx ◦ y
gxgy holds for a set X with a binary operation ◦ under an additional assumption.
2. Main Result
Theorem 2.1. Let ε ≥ 0 and X a set with a binary operation ◦ such that, for each x, y, z ∈ X,either
x ◦ y
◦ z
x ◦ z
◦ y, or z ◦
x ◦ y
x ◦
z ◦ y
. 2.1
If f : X → C satisfies
f
x ◦ y
− f
x
f
y
≤ ε
∀x, y ∈ X
, 2.2
then fx ◦ yfxfy
for all x, y ∈ X,or|fx|≤1
√
1 4ε /2 for all x ∈ X. In the latter
case, the constant 1
√
1 4ε /2 is the best possible.
Proof. Let f : X → C be a functional satisfying 2.2. Suppose that f is bounded. There exists
a constant C<∞ such that |fx|≤C for all x ∈ X.SetM sup
x∈X
|fx| < ∞.By2.2,we
have, for each x ∈ X, |fx ◦ x − fx
2
|≤ε, and therefore
f
x
2
≤ ε
f
x ◦ x
≤ ε M. 2.3
Thus, M
2
≤ ε M. Now it is easy to see that M ≤ 1
√
1 4ε /2. Consequently, if f is
bounded, then |fx|≤1
√
1 4ε /2 for all x ∈ X. The constant 1
√
1 4ε /2isthe
best possible since gx1
√
1 4ε /2forx ∈ X satisfies gxgy − gx ◦yε for each
x, y ∈ X. It should be mentioned that the above proof is essentially due to P.
ˇ
Semrl 5, Proof
of Theorem 2.1 and Proposition 2.2cf. 6,Proposition5.5.
Suppose that f : X → C is an unbounded functional satisfying the inequality 2.2.
Since f is unbounded, there exists a sequence {z
k
}
k∈N
⊂ X such that lim
k →∞
|fz
k
| ∞. Take
x, y ∈ X arbitrarily. Set
N
1
k ∈ N :
x ◦ y
◦ z
k
x ◦ z
k
◦ y
,
N
2
k ∈ N : z
k
◦
x ◦ y
x ◦
z
k
◦ y
.
2.4
By 2.1, N N
1
∪ N
2
. Thus either N
1
or N
2
is an infinite subset of N. First we consider the
case when N
1
is infinite. Take k
1
∈ N
1
arbitrarily. Choose k
2
∈ N
1
with k
1
<k
2
. Since N
1
is
Journal of Inequalities and Applications 3
assumed to be infinite, for each m>2 there exists k
m
∈ N
1
such that k
m−1
<k
m
. Then {z
k
m
}
m∈N
is a subsequence of {z
k
}
k∈N
with k
m
∈ N
1
for every m ∈ N. By the choice of {z
k
}
k∈N
, we have
lim
m →∞
f
z
k
m
∞. 2.5
Thus we may and do assume that fz
k
m
/
0 for every m ∈ N.By2.2 we have, for each
w ∈ X and m ∈ N, |fw ◦ z
k
m
− fwfz
k
m
|≤ε. According to 2.5, we have
f
w ◦ z
k
m
f
z
k
m
− f
w
≤
ε
f
z
k
m
−→ 0asm →∞. 2.6
Consequently, we have, for each w ∈ X,
f
w
lim
m →∞
f
w ◦ z
k
m
f
z
k
m
. 2.7
Since k
m
∈ N
1
, we have x ◦y ◦ z
k
m
x ◦z
k
m
◦ y for every m ∈ N. Applying 2.7, we have
f
x ◦ y
lim
m →∞
f
x ◦ y
◦ z
k
m
f
z
k
m
lim
m →∞
f
x ◦ z
k
m
◦ y
f
z
k
m
lim
m →∞
f
x ◦ z
k
m
◦ y
− f
x ◦ z
k
m
f
y
f
z
k
m
lim
m →∞
f
x ◦ z
k
m
f
y
f
z
k
m
.
2.8
By 2.2 and 2.5, we have
lim
m →∞
f
x ◦ z
k
m
◦ y
− f
x ◦ z
k
m
f
y
f
z
k
m
≤ lim
m →∞
ε
f
z
k
m
0. 2.9
Consequently, we have by 2.8 and 2.7
f
x ◦ y
lim
m →∞
f
x ◦ z
k
m
f
y
f
z
k
m
lim
m →∞
f
x ◦ z
k
m
f
z
k
m
f
y
f
x
f
y
. 2.10
Next we consider the case when N
2
is infinite. By a quite similar argument as in the
case when N
1
is infinite, we see that there exists a subsequence {z
k
n
}
n∈N
⊂{z
k
}
k∈N
such that
k
n
∈ N
2
for every n ∈ N. Then
lim
n →∞
f
z
k
n
∞. 2.11
4 Journal of Inequalities and Applications
In the same way as in the proof of 2.7, we have
f
w
lim
n →∞
f
z
k
n
◦ w
f
z
k
n
, 2.12
for every w ∈ X. According to 2.2 and 2.11, we have
lim
n →∞
f
x ◦
z
k
n
◦ y
− f
x
f
z
k
n
◦ y
f
z
k
n
≤ lim
n →∞
ε
f
z
k
n
0. 2.13
Since z
k
n
◦ x ◦ yx ◦ z
k
n
◦ y for every n ∈ N, 2.11 and 2.12 show that
f
x ◦ y
lim
n →∞
f
z
k
n
◦
x ◦ y
f
z
k
n
lim
n →∞
f
x ◦
z
k
n
◦ y
f
z
k
n
lim
n →∞
f
x ◦
z
k
n
◦ y
− f
x
f
z
k
n
◦ y
f
z
k
n
lim
n →∞
f
x
f
z
k
n
◦ y
f
z
k
n
lim
n →∞
f
x
f
z
k
n
◦ y
f
z
k
n
f
x
lim
n →∞
f
z
k
n
◦ y
f
z
k
n
f
x
f
y
.
2.14
Consequently, if f is unbounded, then fx ◦ yfxfy for all x, y ∈ X.
Remark 2.2. Let φ be a mapping from a commutative semigroup X into itself. We define the
binary operation ◦ by x ◦ y xφy for each x, y ∈ X. Then ◦ satisfies 2.1 since
x ◦ y
◦ z xφ
y
φ
z
xφ
z
φ
y
x ◦ z
◦ y, 2.15
for all x, y, z ∈ X. Therefore, Theorem 2.1 is a generalization of Najdecki 4, Theorem 1 and
Baker 3, Theorem 1.
Remark 2.3. Let X be a set, and f : X → C. Suppose that X has a binary operation ◦ such that,
for each x, y, z ∈ X, either
f
x ◦ y
◦ z
f
x ◦ z
◦ y
, or f
z ◦
x ◦ y
f
x ◦
z ◦ y
. 2.16
Journal of Inequalities and Applications 5
If f satisfies 2.2 for some ε ≥ 0, then by quite similar arguments to the proof of Theorem 2.1,
we can prove that fx ◦ yfxfy for all x, y ∈ X,or|fx|≤1
√
1 4ε /2 for all
x ∈ X.Thus,Theorem 2.1 is still true under the weaker condition 2.16 instead of 2.2.This
was pointed out by the referee of this paper. The condition 2.16 is related to that introduced
by Kannappan 7.
Example 2.4. Let ϕ and ψ be mappings from a semigroup X into itself with the following
properties.
a ϕxyϕxϕy for every x, y ∈ X.
b ψX ⊂{x ∈ X : ϕxx}.
c ψxψyψyψx for every x, y ∈
X.
If we define x ◦y ϕxψy for each x, y ∈ X, then we have x ◦y ◦z x ◦z ◦y for every
x, y, z ∈ X. In fact, if x, y, z ∈ X, then we have
x ◦ y
◦ z ϕ
x ◦ y
ψ
z
ϕ
ϕ
x
ψ
y
ψ
z
by a
ϕ
2
x
ϕ
ψ
y
ψ
z
by b
ϕ
2
x
ψ
y
ψ
z
by c
ϕ
2
x
ψ
z
ψ
y
by b
ϕ
2
x
ϕ
ψ
z
ψ
y
by a
ϕ
ϕ
x
ψ
z
ψ
y
ϕ
x ◦ z
ψ
y
x ◦ z
◦ y
2.17
as claimed.
Let ϕ be a ring homomorphism from C into itself, that is, ϕz wϕzϕw and
ϕzwϕzϕw for each z, w ∈ C. It is well known that there exist infinitely many such
homomorphisms on C cf. 8, 9.Ifϕ is not identically 0, then we see that ϕqq for every
q ∈ Q, the field of all rational real numbers. Thus, if we consider the case when X C
, ϕ a
nonzero ring homomorphism, and ψ : X → Q, then X, ϕ, ψ satisfies the conditions a, b,
and c.
If we define x ∗y y ◦ x for each x, y ∈ X, then z ∗ x ∗ yx ∗z ∗ y holds for every
x, y, z ∈ X. In fact,
z ∗
x ∗ y
x ∗ y
◦ z
y ◦ x
◦ z
y ◦ z
◦ x x ∗
z ∗ y
. 2.18
6 Journal of Inequalities and Applications
Example 2.5. Let X C ×{0, 1}, and, let ϕ, ψ : C → C. We define the binary operation ◦ by
x, a
◦
y, b
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
xψ
y
, 0
, if a b 0,
ϕ
x
y, 1
, if a b 1,
0, 0
, if a
/
b,
2.19
for each x, a, y, b ∈ X. Then ◦ satisfies the condition 2.1.Infact,let
x, a, y, b, z, c
∈ X.
a If a b c 0, then we have
x, a
◦
y, b
◦
z, c
xψ
y
ψ
z
, 0
x, a
◦
z, c
◦
y, b
. 2.20
b If a b c 1, then
z, c
◦
x, a
◦
y, b
ϕ
z
ϕ
x
y, 1
x, a
◦
z, c
◦
y, b
.
2.21
c If a b 0andc 1, then
x, a
◦
y, b
◦
z, c
0, 0
x, a
◦
z, c
◦
y, b
. 2.22
d If a b 1andc 0, then
z, c
◦
x, a
◦
y, b
0, 0
x, a
◦
z, c
◦
y, b
,
x, a
◦
y, b
◦
z, c
0, 0
x, a
◦
z, c
◦
y, b
.
2.23
e If a
/
b, then we have
x, a
◦
y, b
◦
z, c
0, 0
x, a
◦
z, c
◦
y, b
. 2.24
Therefore, ◦ satisfies the condition 2.1. On the other hand, if a b c 0, then
z, c
◦
x, a
◦
y, b
zψ
xψ
y
, 0
,
x, a
◦
z, c
◦
y, b
xψ
zψ
y
, 0
.
2.25
Thus, z, c ◦ x, a ◦ y, b
/
x, a ◦ z, c ◦ y, b in general. In the same way, we see that
if a b c 1, then x, a ◦ y, b ◦ z, cx, a ◦ z, c ◦ y, b
need not to be true.
Journal of Inequalities and Applications 7
Acknowledgments
The authors would like to thank the referees for valuable suggestions and comments to
improve the manuscript. The first and fourth authors were partly supported by the Grant-
in-Aid for Scientific Research.
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